Calculus Word Problem

Hi, everyone. I'm new to the site and I need help setting up a calculus problem. Here it is:

The stadium vending company finds that sales of hot dogs average 34000 hot dogs per game when the hot dogs sell for $2.50 each. For each 50 cent increase in the price, the sales per game drop by 5000 hot dogs. What price per hot dog should the vending company charge to realize the maximum revenue?

let p be the price per hot dog
let q be the quantity of hot dogs sold
let R be the revenue

I think it should be very clear that

but this is a function of two variables. We have a constraint, however. What you should do is eliminate q from the expression for R. This should be easy because they told you that there is a linear relationship between q and p (you know a point and the slope, so you know every point on the line).

For the R function, after you make your substitution, you will get a 2nd degree polynomial.

Hi, everyone. I'm new to the site and I need help setting up a calculus problem. Here it is:

The stadium vending company finds that sales of hot dogs average 34000 hot dogs per game when the hot dogs sell for $2.50 each. For each 50 cent increase in the price, the sales per game drop by 5000 hot dogs. What price per hot dog should the vending company charge to realize the maximum revenue?

Thanks in advance.

let x be the quantity of hot dogs sold
let p(x) be the price function
let R(x) be the revenue function

The stadium vending company finds that sales of hot dogs average 34000 hot dogs per game when the hot dogs sell for $2.50 each so p(x) has the ordered pair (34000, 2.50)

For each 50 cent increase in the price, the sales per game drop by 5000 hot dogs so p(x) has slope .50/-5000

So p(x) - 2.50 = -.50/5000 (x - 34000)

Solve for p(x) then find R(x)=xp(x)

You can find the value of x that gives the maximum revenue by finding the derivative of R(x), setting it equal to zero and solving for x. If the derivative of R(x) changes from positive to negative at that value of x, you know that R(x) is maximum at that value of x.

let x be the quantity of hot dogs sold
let p(x) be the price function
let R(x) be the revenue function

The stadium vending company finds that sales of hot dogs average 34000 hot dogs per game when the hot dogs sell for $2.50 each so p(x) has the ordered pair (34000, 2.50)

For each 50 cent increase in the price, the sales per game drop by 5000 hot dogs so p(x) has slope .50/-5000

So p(x) - 2.50 = -.50/5000 (x - 34000)

Solve for p(x) then find R(x)=xp(x)

You can find the value of x that gives the maximum revenue by finding the derivative of R(x), setting it equal to zero and solving for x. If the derivative of R(x) changes from positive to negative at that value of x, you know that R(x) is maximum at that value of x.